Problem 15 - University Physics with Modern Physics

Question

How many excess electrons must be added to an isolated spherical conductor 26.0 cm in diameter to produce an electric field of magnitude 1150 N/C just outside the surface?

Step-by-Step Solution

Answer

About 1.34 x 10^10 excess electrons are needed.

Step 1: Understand the given variables

You are given that the diameter of the spherical conductor is 26.0 cm, and you need to find the number of excess electrons required to produce an electric field of 1150 N/C just outside the surface. This involves relating charge, electric field, and the geometry of the sphere.

Step 2: Calculate the radius of the sphere

The radius of the sphere is half of its diameter. Thus, \[ r = \frac{26.0 \text{ cm}}{2} = 13.0 \text{ cm} = 0.13 \text{ m}. \]

Step 3: Apply the formula for electric field due to a charged sphere

The electric field just outside a charged sphere is given by \[ E = \frac{kQ}{r^2}, \] where \( E \) is the electric field, \( k = 8.99 \times 10^9 \text{ N}\,\text{m}^2/\text{C}^2 \) is the Coulomb's constant, \( Q \) is the charge on the sphere, and \( r \) is the radius.

Step 4: Solve for the charge \( Q \)

Rearrange the formula to solve for \( Q \): \[ Q = \frac{Er^2}{k}. \]Substitute the values of \( E = 1150 \text{ N/C} \), \( r = 0.13 \text{ m} \), and \( k = 8.99 \times 10^9 \text{ N}\,\text{m}^2/\text{C}^2 \):\[ Q = \frac{(1150)(0.13)^2}{8.99 \times 10^9} \approx 2.15 \times 10^{-9} \text{ C}. \]

Step 5: Find the number of excess electrons

The charge of one electron is \( e = 1.6 \times 10^{-19} \text{ C} \). To find the number of excess electrons \( n \), use the formula:\[ n = \frac{Q}{e}. \]Substitute \( Q = 2.15 \times 10^{-9} \text{ C} \):\[ n = \frac{2.15 \times 10^{-9}}{1.6 \times 10^{-19}} \approx 1.34 \times 10^{10}. \]

Key Concepts

Coulomb's Law

Coulomb's Law is fundamental in understanding the interactions between charged objects. It describes how the electric force between two charged objects depends on their charges and the distance between them.
Coulomb's Law formula is given by:\[F = k \frac{|q_1 q_2|}{r^2}\]where:

• \(F\) is the force between the charges.
• \(q_1\) and \(q_2\) are the magnitudes of the charges.
• \(r\) is the distance between the charges.
• \(k\) is Coulomb's constant (\(8.99 \times 10^9 \text{ N}\,\text{m}^2/\text{C}^2\)).

When dealing with spheres, things become interesting as the charges can be imagined to be concentrated at the center. This makes calculating the field outside a charged sphere much simpler, as inside the electric field, any charge can be disregarded if calculated at a point beyond the sphere's surface.
An important aspect to remember is that, just like gravity, the electric force is inversely proportional to the square of the distance. This principle is crucial for scenarios like charged spheres where symmetry simplifies calculations.

Excess Electrons

Excess electrons refer to additional electrons given to an object, imparting it a negative charge. Electrons are negatively charged subatomic particles, and when you add them to a neutral object, they increase its charge.
In this context, the number of excess electrons determines the overall charge on the sphere. It follows the formula:\[Q = n \cdot e\]where:

• \(Q\) is the total charge.
• \(n\) is the number of excess electrons.
• \(e\) is the charge of a single electron (\(1.6 \times 10^{-19} \text{ C}\)).

By rearranging the formula, you can calculate how many electrons you need to achieve a specific charge. When dealing with charged objects, knowing the charge of a single electron helps translate the macroscopic quantities into a microscopic count for electrons.
This concept helps in understanding how large charges are built up from elementary particles. Knowing the number of excess electrons provides insight into the charge intensity and its corresponding electrostatic forces.

Charged Sphere

A charged sphere is a common model used in physics to simplify the analysis of electric fields. A sphere's symmetric shape allows electric charge to distribute evenly across its surface when it is an ideal conductor. This uniform distribution simplifies the calculations of electric fields around the sphere.
For a sphere conducting electricity, the electric field outside of it is defined by:\[E = \frac{kQ}{r^2}\]Here:

• \(E\) represents the electric field outside the sphere.
• \(Q\) is the charge on the sphere.
• \(r\) is the distance from the center of the sphere to where you're measuring the field.
• \(k\) is Coulomb's constant.

This formulation shows that outside a charged sphere, the electric field behaves as though all of the sphere's charge is concentrated at its center. This simplification is a powerful tool for calculating field strengths without complex geometry.
Understanding charged spheres lays the groundwork for advancing studies in electromagnetism and electronics, where spheres or spherical segments are regularly encountered.